Complex Locus (2 Viewers)

[qwe]

Need help with a locus question, question below

Find the locus of "modulus(z-3) + modulus(z+3) =5"

Looks like an ellipse but when i let z=x+iy it comes out as a hyperbola........

Any ideas people?

 

[khorne]

Nevermind

 

[study-freak]

|z-3| + |z+3| = 5
|x-3 + iy| + |x+3 + iy| = 5
(x-3)^2 + y^2 + (x+3)^2 + y^2 = 25
x^2 + 18 + 2y^2 = 25
x^2+ 2y^2 = 7 (ellipse)

???

|x-3 + iy| + |x+3 + iy| = 5
(x-3)^2 + y^2 + (x+3)^2 + y^2 +2 sqrt[{(x-3)^2 + y^2)}{(x+3)^2 + y^2)}] = 25

 

[khorne]

My bad.

 

[Lukybear]

Wait... wats the problem?

 

[Trebla]

Need help with a locus question, question below

Find the locus of "modulus(z-3) + modulus(z+3) =5"

Looks like an ellipse but when i let z=x+iy it comes out as a hyperbola........

Any ideas people?

There's no way that could be an ellipse.
The geometric interpretation for an ellipse uses:
PS + PS' = 2a
The problem lies in the fact the SS' = 6 (draw a diagram to see this), so the sum of two sides (PS and PS') of triangle PSS' is LESS than the third side which is a violation of the triangle inequality!

Hence there is NO valid locus that satisfies the condition in the question...

Algebraically, the flaw is MUCH harder to spot.



Now consider, if the question was instead written like:


We end up with the SAME equation, either way because the step where we square both sides eliminates the information of the negative sign in the original question! In the second case, the geometric interpretation is correct using PS - PS' = 2a (provided PS > PS') and the triangle inequality is not violated.

The flaw in the original part lies in the bits in red and blue.

For the part in red, the LHS of the equality is non-negative so the RHS must obviously also be non-negative which leads to the condition:



Note that squaring both sides here is OKAY because both sides are positive.

For the part the blue, again the LHS of the equality is non-negative so the RHS must obviously also be non-negative which leads to the condition:



So for the final answer to be valid, BOTH the following conditions must be satisfied:



This can easily be shown graphically.

Now consider the final answer obtained which was



If you consider the sketch or the natural domain (x ≥ 2.5 and x ≤ - 2.5) of this hyperbola and compare that with the sketch of the two restrictions mentioned above, you will notice that NEITHER can be simultaneously satisfied i.e. the hyperbola does not exist under the given restrictions.

This means that there is NO valid solution for the locus.

Take home message: take care when squaring both sides!!!!!

 

[qwe]

Wow nice job on the algebraic explanation Trebla, and thank you for answering the question.

I guess I should have seen the triangle inequality violation.

 

[khorne]

That is why I think doing it geometrically preserves is better, as you preserve this information.

 

[adomad]

for this question, find the locus of z if Re(z) =|z|. can't you just say that it is only logical if the locus is the Re(z) axis where x>=0??

and how the fiffie cakes do you do this one

find the value of k if arg(z-2)=k arg(z^2-2z)?

 

[addikaye03]

for this question, find the locus of z if Re(z) =|z|. can't you just say that it is only logical if the locus is the Re(z) axis where x>=0??

and how the fiffie cakes do you do this one

find the value of k if arg(z-2)=k arg(z^2-2z)?

arg(z-2)=k arg(z^2-2z)

z-2=(z^2-2z)^k

z-2=z^k(z-2)^k

(z^k) (z-2)^(k-1)=1

Got it from there?

 

[study-freak]

arg(z-2)=k arg(z^2-2z)

z-2=(z^2-2z)^k

Just because their arguments are equal, it doesn't mean that modulus of the expressions inside the brackets are also equal.

 

[adomad]

LOL. i forgot the read the start of the question. just before i was going to rage quit, i scrolled up


it states that |z-2| = 2 and 0< arg(z) <90 degrees

i got it now. thanks guys.

 

[GUSSSSSSSSSSSSS]

LOL. i forgot the read the start of the question. just before i was going to rage quit, i scrolled up


it states that |z-2| = 2 and 0< arg(z) <90 degrees

i got it now. thanks guys.

hahahahaha well saying "that helps" wud be a huuuuuuuuuuuuuuuge UNDERSTATEMENT

lol

 

[gurmies]

k = 2/3

 

[shaon0]

k = 2/3

Damn, i got k=1/3.

 

[adomad]

i also got 2/3

 

[gurmies]

Could somebody post an algebraic solution, mine was geometric?

 

[shaon0]

arg(z-2)=k.arg(z^2-2z)
If mod(z-2)=2, then;
arg(2)=k.arg(2z) [idk if this is correct]
But, mod(z-2)=2
ie. z-2=2 or z-2=-2
Thus, z=4 or 0 but arg(0) is undefined ie. z=4

arg(2)=k.arg(8)
arg(2)=3k.arg(2)
k=1/3

This solution is probably incorrect as I'm rusty with 4u
Gurmies, how did you do it geometrically?

 

[Jacob1991]

arg(z-2)=k arg(z^2-2z)

z-2=(z^2-2z)^k

z-2=z^k(z-2)^k

(z^k) (z-2)^(k-1)=1

Got it from there?

if this is the question
'find k if arg(z-2)=k arg(z^2-2z)' and there are no more conditions, then k varies depending on z doesnt it? at least that's what Simon and I think... but our 4U knowledge is getting rusty so yeh...

 

[study-freak]

Could somebody post an algebraic solution, mine was geometric?

I think geometric way is the best way.